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<math>X(t) = 6\frac{e^{j2t}+e^{-j2t}}{2} + 8\frac{e^{j4t}-e^{-j4t}}{2}\,</math><br> | <math>X(t) = 6\frac{e^{j2t}+e^{-j2t}}{2} + 8\frac{e^{j4t}-e^{-j4t}}{2}\,</math><br> | ||
<math>X(t) = 3e^{j2t}+3e^{-j2t} + 4e^{j4t}-4e^{-j4t}\,</math><br> | <math>X(t) = 3e^{j2t}+3e^{-j2t} + 4e^{j4t}-4e^{-j4t}\,</math><br> | ||
| + | With this expression we can conclude: | ||
| + | <math>a_1 = 3\,</math><br> | ||
| + | <math>a_{-1} = 3\,</math><br> | ||
| + | <math>a_2 = 4\,</math><br> | ||
| + | <math>a_{-2} = 4\,</math><br> | ||
| + | |||
| + | To calculate <math>a_k\,</math>, we use this equation: | ||
| + | <math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\,</math><br> | ||
Revision as of 19:02, 21 September 2008
CT Signal:
$ X(t) = 6\cos(2t) + 8\sin(4t)\, $
$ X(t) = 6\frac{e^{j2t}+e^{-j2t}}{2} + 8\frac{e^{j4t}-e^{-j4t}}{2}\, $
$ X(t) = 3e^{j2t}+3e^{-j2t} + 4e^{j4t}-4e^{-j4t}\, $
With this expression we can conclude:
$ a_1 = 3\, $
$ a_{-1} = 3\, $
$ a_2 = 4\, $
$ a_{-2} = 4\, $
To calculate $ a_k\, $, we use this equation:
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\, $
